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  • hdu 5672 Strings 模拟

    Problem Description
    There is a string S.S only contain lower case English character.(10length(S)1,000,000)
    How many substrings there are that contain at least k(1k26) distinct characters?
     

    Input
    There are multiple test cases. The first line of input contains an integerT(1T10) indicating the number of test cases. For each test case:

    The first line contains string S.
    The second line contains a integer k(1k26).
     

    Output
    For each test case, output the number of substrings that contain at leastk dictinct characters.
     

    Sample Input
    2 abcabcabca 4 abcabcabcabc 3
     

    Sample Output
    0 55
     
    题目:给一个字符串,问有多少个子串至少含有k个不同的字母?

    分析:从头开始扫一遍,如果遇见第一次出现过的,就计数cnt++,直到cnt==k,那么就从那个子串的开头开始删去字符,根据删除的字符出现的次数,做出相应的判断。

    如果用map判断是否出现过会超时。

    #include<cstring>
    #include<string>
    #include<cstdio>
    #include<algorithm>
    #include<queue>
    #include<map>
    #include<set>
    #include<vector>
    #include<stack>
    using namespace std;
    typedef long long ll;
    const int N=1e6+9;
    char s[N];
    int k;
    int vis[300];
    int main()
    {
        //freopen("f.txt","r",stdin);
        int T;scanf("%d",&T);
        while(T--){
            scanf("%s%d",s,&k);
            int slen=strlen(s);
            if(k==1){
                printf("%I64d
    ",(ll)slen*(slen+1)/2);continue;
            }
            int head=0;
            ll ans=0;
            int cnt=0;
            memset(vis,0,sizeof(vis));
            for(int i=0;i<slen;i++){
                if(vis[s[i]]>0){
                    vis[s[i]]++;
                    continue;
                }
                vis[s[i]]=1;
                cnt++;
                if(cnt==k){
                    while(head<i){
                        ans+=slen-i; 
                        if(vis[s[head]]==1){
                            vis[s[head]]=0;
                            head++;
                            cnt--;
                            break;
                        }
                        else{
                            vis[s[head]]--;head++;
                        }
                    }
                }
            }
            printf("%I64d
    ",ans);
        }
        return 0;
    }
    



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  • 原文地址:https://www.cnblogs.com/01world/p/5651223.html
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